Chapter 21: Problem 37
(II) A very thin line of charge lies along the \(x\) axis from \(x = - \infty\) to \(x = + \infty .\) Another similar line of charge lies along the \(y\) axis from \(y = - \infty\) to \(y = + \infty .\) Both lines have a uniform charge per length \(\lambda .\) Determine the resulting electric field magnitude and direction (relative to the \(x\) axis) at a point \(( x , y )\) in the first quadrant of the \(x y\) plane.
Short Answer
Step by step solution
Understanding the Charge Distribution
Electric Field due to a Uniform Line of Charge
Electric Field due to the Line along the x-axis
Electric Field due to the Line along the y-axis
Resultant Electric Field Vector
Direction of the Resultant Electric Field
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Line Charge
Charge Density
Vector Addition
- The electric field from the x-axis line is perpendicular to the x-axis, acting along the y-direction.
- The electric field from the y-axis line is perpendicular to the y-axis, acting along the x-direction.
Uniform Charge Distribution
- The infinite nature of the lines ensures that every small segment contributes equally, creating a uniform field along specific vectors.
- This uniformity ensures that at any given point in the plane, the contributions from each charge are symmetric and evenly distributed.